Php web service to get data in json format - php

Question

I have 2 database tables
main_menu which has 3 columns id, item_name, img
sub_menu which has columns id, sub_item_name, item_name, add_price, img
Now I am trying to get all sub menu under master menu when user clicks on master menu
eg.
select * from sub_menu where item_name="Cafe"
it should display all sub menu under Cafe.
And here is the web service I wrote to do that and get data in json format
<?php
$response = array();
// include db connect class
require_once __DIR__ . '/db_connect.php';
// connecting to db
$db = new DB_CONNECT();
// check for post data
if (isset($_GET["item_name"])) {
// get a product from products table
$result = mysql_query("SELECT * FROM sub_menu WHERE item_name ='".$_GET["item_name"]."'");
echo $result;
if (!empty($result)) {
// check for empty result
if (mysql_num_rows($result) > 0) {
$result = mysql_fetch_array($result);
$sub_menu = array();
$sub_menu["id"] = $result["id"];
$sub_menu["item_name"] = $result["item_name"];
$sub_menu["sub_item_name"] = $result["sub_item_name"];
$sub_menu["add_price"] = $result["add_price"];
$sub_menu["img"] = $result["img"];
// success
$response["success"] = 1;
// user node
$response["sub_menu"] = array();
array_push($response["sub_menu"], $sub_menu);
// echoing JSON response
echo json_encode($response);
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No sub menu found";
// echo no users JSON
echo json_encode($response);
}
} else {
// no product found
$response["success"] = 0;
$response["message"] = "No menu found";
// echo no users JSON
echo json_encode($response);
}
} else {
// required field is missing
$response["success"] = 0;
$response["message"] = "Required field(s) is missing";
// echoing JSON response
echo json_encode($response);
}
?>
Now the problem is I am only getting first data only, it should display all data
Query is working fine in mysql, I checked that. The problem is somewhere in my php service.

Answer 1

You have to call mysql_fetch_array() in a loop, and add all the rows to an array. You can just copy the whole row of results in a single assignment, since you were using the same field names in your JSON as in the database.
if (mysql_num_rows($result) > 0) {
$submenu = array();
while ($result = mysql_fetch_array($result);
$submenu[] = $result;
}
$response["success"] = 1;
// user node
$response["sub_menu"] = $all_results;
// echoing JSON response
echo json_encode($response);